Please help...The area of a rectangular room is given by the trinomial x^2+4x-12. What are the possible dimensions of the rectangle? Use factoring.

A. (x-6) and (x-2)

B. (x+6) and (x+2)
C. (x-6) and (x+2)
D. (x+6) and(x-2)
The correct answer (for the test I took) is D.

(x-2)(x +6)

thanks

Oh boy, let's solve this puzzle using some clown logic, I mean factoring! So we have a trinomial, x^2 + 4x - 12, which represents the area of our rectangular room.

First, we need to find two numbers that multiply to give -12 (the constant term), and add up to 4 (the coefficient of the x-term). How about 6 and -2? They both multiply to -12 and add up to 4.

Now, let's split the middle term into two parts using these numbers. We can rewrite the trinomial as x^2 + 6x - 2x - 12.

Now, let's factor by grouping. We can factor out an x from the first two terms and factor out a -2 from the last two terms. This gives us x(x + 6) - 2(x + 6).

Now, we have a common binomial factor of (x + 6). We can factor that out, which gives us (x - 2)(x + 6).

So, the possible dimensions of the rectangle are (x - 2) and (x + 6). Clown math at its finest!

To find the possible dimensions of the rectangle, we need to factor the given trinomial x^2 + 4x - 12.

Step 1: Start by looking for two numbers that multiply to give the constant term (-12) and add up to give the coefficient of the middle term (+4).

The pair of numbers that satisfy both of these conditions is 6 and -2, since 6 * (-2) = -12 and 6 + (-2) = 4.

Step 2: Rewrite the middle term (+4x) using these two numbers:
x^2 + 6x - 2x - 12

Step 3: Group the terms and factor them:
(x^2 + 6x) + (-2x - 12)
x(x + 6) - 2(x + 6)

Step 4: Notice that we have a common binomial factor, (x + 6), which we can factor out:
(x - 2)(x + 6)

So, the factored form of the trinomial x^2 + 4x - 12 is (x - 2)(x + 6).

Step 5: Equate each factor to zero and solve for x:
x - 2 = 0 or x + 6 = 0

Solving these equations, we find two possible values for x:
x = 2 or x = -6

Step 6: Interpret the solutions in the context of the problem. Since the dimensions of a rectangle cannot be negative, we discard x = -6. Therefore, the possible dimension of the rectangle is x = 2.

Hence, the possible dimensions of the rectangle are 2 units by (2 + 6) = 2 units by 8 units.